Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s-1. Take 'g' constant with a value 10 m s-2. The work done by the (i) gravitational force and the (ii) resistive force of air is
- (i) 1.25 J (ii) -8.25 J
- (i) 100 J (ii) 8.75 J
- (i) 10 J (ii) -8.75 J
- (i) -10 J (ii) -8.25 J
Answer
(3) (i) 10 J (ii) -8.75 J
Solution
(i) Wg = mgh = (10-3)(10)(103) = 10 J
Kinetic energy on reaching the ground = 12mv2 = 12(10-3)(50)2 = 1.25 J
(ii) By work–energy theorem, Wg + Wair = K
Wair = 1.25 - 10
= -8.75 J
Answer: (3)